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Checkpoint: Assess Your Understanding, pages 197–199
3.1
1. Multiple Choice Which expression is the factored form of
3x2 + 11x - 4?
A. (3x + 4)(x - 1)
B. (3x + 1)(x - 4)
C. (3x - 1)(x + 4)
D. (3x - 4)(x + 1)
2. Factor.
a) 36x2 - 49y2
(6x)2 (7y)2
(6x 7y)(6x 7y)
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b) 0.5x2 - 3.5x + 5
0.5(x2 7x 10)
0.5(x 2)(x 5)
Chapter 3: Solving Quadratic Equations—Checkpoint—Solutions
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c) 10x2 + 29x - 21
1
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1
d) 5x2 ⫺ 180 y 2
1 2
yb
36
10x2 35x 6x 21
ax2 5x(2x 7) 3(2x 7)
ax yb ax yb
1
5
1
5
1
6
1
6
(5x 3)(2x 7)
3. Factor.
a) (7x + 4)2 - (3y - 2)2
[(7x 4) (3y 2)][(7x 4) (3y 2)]
[7x 4 3y 2][7x 4 3y 2]
(7x 3y 2)(7x 3y 6)
b) 3(2x - 1)2 + 14(2x - 1) + 8
[3(2x 1) 2][(2x 1) 4]
(6x 3 2)(2x 1 4)
(6x 1)(2x 3)
4. Determine whether 2x - 5 is a factor of each polynomial.
a) 10x2 + 23x - 5
Write the trinomial as:
(2x 5)(5x b)
10x2 (2b 25)x 5b
Equate constant terms.
5b 5, so b 1
Check:
(2x 5)(5x 1)
10x2 23x 5
So, 2x 5 is not a factor.
b) 6x2 - 17x + 5
Write the trinomial as:
(2x 5)(3x b)
6x2 (2b 15)x 5b
Equate constant terms.
5b 5, so b 1
Check:
(2x 5)(3x 1)
6x2 17x 5
So, 2x 5 is a factor.
3.2
5. Multiple Choice Which values of x are solutions of 3x2 + 2x = 8?
4
3
A. x = 3, x = -2
B. x = 4, x = -2
2
3
C. x = 3, x = -4
D. x = 2, x = -4
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Chapter 3: Solving Quadratic Equations—Checkpoint—Solutions
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6. Solve by factoring, then verify each solution.
a) x2 - 8x - 33 = 0
b) 8x2 + 22x - 21 = 0
(x 11)(x 3) 0
Either x 11 0,
then x 11;
or x 3 0, then x 3
(2x 7)(4x 3) 0
Either 2x 7 0,
then x 3.5; or
4x 3 0, then x 0.75
7. Solve each equation.
a) (x - 2)(x + 3) = 24
b) 5x2 - 20x = x2 + 8x - 49
x2 x 6 24 0
x2 x 30 0
(x 5)(x 6) 0
Either x 5 0, then x 5;
or x 6 0, then x 6
5x2 20x x2 8x 49 0
4x2 28x 49 0
(2x 7)(2x 7) 0
2x 7 0, then x 3.5
8. Solve each equation, then verify the solution.
a)
√
4x + 3 = x
√
4x x 3
√ 2
( 4x) (x 3)2
b)
4x x2 6x 9
0 x2 10x 9
0 (x 9)(x 1)
Either x 9 0, then x 9;
or x 1 0, then x 1
I used mental math to verify.
x 1; the root is x 9
√
2x - 7 + 5 = x
√
2x 7 x 5
√
( 2x 7)2 (x 5)2
2x 7 x2 10x 25
0 x2 12x 32
0 (x 4)(x 8)
Either x 4 0, then x 4;
or x 8 0, then x 8
I used mental math to verify.
x 4; the root is x 8
9. The diagonal of a rectangle is 17 cm long. The rectangle is 7 cm
longer than it is wide. What are the dimensions of the rectangle?
Let the width of the rectangle be x centimetres.
Then the length of the rectangle, in centimetres, is: x 7
Use the Pythagorean Theorem to write an equation.
x2 (x 7)2 172
2
2
x x 14x 49 289 0
2x2 14x 240 0 Divide each term by 2.
x2 7x 120 0
(x 15)(x 8) 0
Either x 15 0
or
x80
x 15
x8
Since the width cannot be negative, the width is 8 cm, and the length
is: (8 7) cm, or 15 cm
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Chapter 3: Solving Quadratic Equations—Checkpoint—Solutions
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